Решите уравнение:[latex]a) 5x-18 \sqrt{x} -8=0[/latex][latex]b) \sqrt{33-8x} =x[/latex][latex]c) 3x-2 \sqrt{x} -8=0[/latex][latex] d)\sqrt{2x+15} =x[/latex]

[latex]5x-18 \sqrt{x} -8=0 \\\ D_1=9^2+5\cdot8=121 \\\ \sqrt{x} = \frac{9+11}{5} =4 \\\ \sqrt{x} \neq \frac{9-11}{5} =-0.4 \\\ x=4^2=16[/latex] Ответ: 16 [latex]\sqrt{33-8x} =x \\\ \left \{ {{33-8x=x^2} \atop {x \geq 0}} \right. \\\ \left \{ {{x^2+8x-33=0} \atop {x \geq 0}} \right. \\\ \left \{ {{D_1=4^2+33=49} \atop {x \geq 0}} \right. \\\ \left \{ {{x=-4\pm7} \atop {x \geq 0}} \right. \\\ \left \{ {{x=-11; x=3} \atop {x \geq 0}} \right. \\\ x=3[/latex] Ответ: 3 [latex]3x-2 \sqrt{x} -8=0 \\\ D_1=1+3\cdot8=25 \\\ \sqrt{x} = \frac{1+5}{3} =2 \\\ \sqrt{x} \neq \frac{1-5}{3} =- \frac{4}{3} \\\ x=2^2=4[/latex] Ответ: 4 [latex]\sqrt{2x+15} =x \\\ \left \{ {{2x+15=x^2} \atop {x \geq 0}} \right. \\\ \left \{ {{x^2-2x-15=0} \atop {x \geq 0}} \right. \\\ \left \{ {{D_1=1+15=16} \atop {x \geq 0}} \right. \\\ \left \{ {{x=1\pm4} \atop {x \geq 0}} \right. \\\ \left \{ {{x=-3; x=5} \atop {x \geq 0}} \right. \\\ x=5[/latex] Ответ: 5